Question

tinh gia tri bieu thuc:

a) M = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{99x100}\)

b)N = \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{97x99}\)

\(x\)la dau nhan

gâp ạ

Answer 1

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow M=1-\frac{1}{100}\)

\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)

\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)

Answer 2

\(a,M=1-\frac{1}{100}=\frac{99}{100}\)

\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)

                  \(=1-\frac{1}{99}=\frac{98}{99}\)

   =>\(N=\frac{98}{99}:2=\frac{49}{99}\)

Answer 3

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(M=1-\frac{1}{100}\)

\(M=\frac{99}{100}\)

Answer 4

a) \(M=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(M=\frac{1}{1}-\frac{1}{100}\)

\(M=\frac{100}{100}-\frac{1}{100}\)

\(M=\frac{99}{100}\)

b) \(N=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\)\(\frac{1}{97\cdot99}\)

\(2N=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(2N=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}\)\(+\frac{1}{5\cdot7}+...+\frac{1}{97\cdot99}\)

\(2N=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(2N=\frac{1}{1}-\frac{1}{99}\)

\(2N=\frac{99}{99}-\frac{1}{99}\)

\(2N=\frac{98}{99}\)

\(N=\frac{98}{99}:2\)

\(N=\frac{49}{99}\)

Answer 5

a) \(M=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{99\cdot100}\)

\(\Leftrightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow M=1-\frac{1}{100}=\frac{99}{100}\)

Vậy \(M=\frac{99}{100}\)